# Math Help - Help with real analysis/advanced calculus

1. ## Help with real analysis/advanced calculus

Hey can anyone help me prove this problem?

Let A and B be a subsets of R and bounded above.
Let InfA=a and InfB=b.
Let C= {xy ; x belongs to A, and y belongs to B}
Prove that InfC=ab

2. ## Re: Help with real analysis/advanced calculus

Originally Posted by salmaster
Let A and B be a subsets of R and bounded above.
Let InfA=a and InfB=b.
Let C= {xy ; x belongs to A, and y belongs to B}
Prove that InfC=ab
Did you mean that to read below?

3. ## Re: Help with real analysis/advanced calculus

This problem isn't true as stated.

1) I think you're intending to assume that A and B are bounded below.

2) Let A = {-1, 1}, B = {-1}. Then a = -1, b = -1, so ab = 1. Also C = {-1, 1}, so inf(C) = -1.
Thus inf(C) is NOT equal to ab.

4. ## Re: Help with real analysis/advanced calculus

Yes I'm sorry I meant bounded below.

5. ## Re: Help with real analysis/advanced calculus

Originally Posted by salmaster
Yes I'm sorry I meant bounded below.
Let $A=\{-1,2\}~\&~B=\{2,-3\}$ then $C=\{-2,3,4,-6\}$
Does the statement hold in that case?

6. ## Re: Help with real analysis/advanced calculus

I don't think that we are supposed to prove it with specific examples of ordered pairs for A, B, and C..

7. ## Re: Help with real analysis/advanced calculus

Originally Posted by salmaster
I don't think that we are supposed to prove it with specific examples of ordered pairs for A, B, and C..
The point is that in reply #5 $\inf(A)=-1,~\inf(B)=-3~\&~\inf(C)=-6$ so that is a counterexample to what you are trying to show.
So the statement you have is false.